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Prin război mondial se înțelege un război care implică mai multe state, mai ales mari puteri ale lumii, un conflict militar la care participă, direct sau indirect, numeroase state ale lumii, extins geografic pe mai multe continente și implicând utilizarea nelimitată a tuturor resurselor strategice disponibile, având drept consecințe schimbări la nivel internațional.

În lumea academică, în sistemul de învățământ, cât și în mass-media, termenul este folosit pentru cele două mari războaie globale purtate între anii 1914-1918 și între anii 1939-1945 (Primul și al Doilea Război Mondial). Au existat antecedente ce corespund caracteristicilor unui război mondial, precum războiul de șapte ani sau războaiele napoleoniene.

Originea termenului
În Dicționarul englez Oxford este citat un pasaj dintr-un ziar scoțian "The People's Journal" dintr-o ediție din anul 1948: " Un război dus între marile puteri este acum în mod necesar un război mondial. "Termenul" război mondial "a fost folosit în 1850 de către Karl Marx și asociatul său, Friedrich Engels, pentru luptele de clasă din Franța. În 1889, a descris un episod din mitologia teutonică ca un "război mondial" (suedez: världskrig), justificând această descriere printr-o linie dintr-un poem epic vechi norvegian, "Völuspá: folcvig fyrst i heimi" ").   Scriitorul german August Wilhelm Otto Niemann a folosit termenul de "război mondial" în titlul romanului anti-britanic" Der Weltkrieg: Deutsche Träume "din 1904, publicat în limba engleză cu titlul "Cucerirea viitoare a Angliei".

În limba engleză, termenul de "primul război mondial" (First World War) a fost utilizat de Charles à Court Repington  ca un titlu pentru jurnalul sau de memorii publicat în 1920.
Termenul de "Primul Război Mondial"  a fost utilizat ulterior în revista Time, termenul de "Al Doilea Război Mondial"  fiind folosit pentru următorul război speculativ.
Pe 4 septembrie, la o zi după ce Franța și Marea Britanie i-au declarat război Germaniei, pe prima pagină a  ziarului danez Kristeligt Dagblad  scria "Cel de-al Doilea Război Mondial a început ieri la ora 11 dimineață".

În limba engleză este folosit termenul  "world war", în franceză este tradus în guerre mondiale, în germană: Weltkrieg, în italiană: guerra mondiale, în spaniolă și portugheză: guerra mundial, iar în rusă: мировая война (mirovaya voyna.)

Istoric
Majoritatea războaielor purtate în era pre-industrială au fost la nivel local, regional, continental și intercontinental. Conflicte militare precum Războiul de 30 de ani sau Războaiele Napoleoniene nu pot fi  catalogate ca războaie globale, ci ca  războaie regionale sau continentale, având în vedere că pentru un război purtat la scară mondială trebuie să fie implicate state de pe mai multe continente sau chiar de pe toată planetă.
De asemenea, chiar dacă s-au implicat puteri europene ca Imperiul Britanic, Regatul Franței, Provinciile Unite și Regatul Spaniei, Războiul de Independență al Statelor Unite ale Americii este un război continental, bătăliile fiind purtate doar pe continentul american.

Sunt istorici care au catalogat Războiul de Șapte Ani ca fiind primul război mondial sau ca războiul mondial 0. Acesta a fost desfășurat în Europa, America de Nord, America de Sud, Africa și Asia, avându-i drept combatanți puteri din două tabere: Imperiul Britanic, Regatul Prusiei, Regatul Portugaliei și Confederația Irocheză vs Regatul Franței, Sfântul Imperiu Roman, Imperiul Rus, Imperiul Spaniol, Imperiul Suedez și Imperiul Mughal. În urmă acestui război au rezultat peste 1 500 000 de victime.

Însuși politicianul britanic Winston Churchill consideră Războiul de Șapte ani drept primul război global.
Cu toate astea, războiul din anii 1914-1918 rămâne în continuare considerat  de către mediul academic și mass-media drept  primul război mondial.

Primul Război Mondial (1914-1918)

Primul Război Mondial a fost declanșat în 1914, fiind primul conflict militar desfășurat la scară globală. Cea de-a două revoluție industrială, accelerarea producției în masă și amploarea globalizării au extins conflictul inițial declanșat la nivel local  între Imperiul Austro-Ungar și Serbia la nivel continental prin intrarea Germaniei, Franței și Marii Britanii și în cele din urmă la nivel mondial prin intrarea Rusiei , Americii și al altor state extra-europene în confruntare.
Războiul a fost desfășurat între două alianțe opuse create anterior -  Puterile Centrale și Antanta, fiecare putere participând la război din anumite motive ce țin de interese proprii, ce țineau fie de dobândire a supremației teritoriale, fie din motive de apărare strategică.  Austro-Ungaria caută să-și consolideze dominația teritorială în peninsula balcanică și Europa centrală, Imperiul Rus caută să-și extindă supremația spre Balcani prin intermediul mișcării panslaviste, Imperiul German concura cu Imperiul Britanic pentru poziția de superputere navală și deținerea coloniilor în Africa, simțindu-se amenințat totodată de tratatul de ajutor reciproc dintre Franța și Imperiul Rus ce îl înconjurau. Imperiul Britanic urmărea să-și asigure supremația globală, iar Franța avea frustrări privind pierderea unor teritorii în față Germaniei în războiul franco-prusac.
America a intrat în război de partea Antantei din cauza agresivității germane ce-i amenință apărarea, iar intervenția acesteia  a dus la încheierea războiului.
Au fost duse lupte în Europa Vestică, Europa estică, Europa balcanică, cât și în Asia, Insulele din Pacific, Oceanul Indian, Oceanul Atlantic, Africa, Orientul Mijlociu, dar și pe coastele continentelor americane, fiind implicate și coloniile imperiilor respective din toată lumea.
Au fost utilizate arme  și inovații militare de ultima generație:  mitraliere, avioane, tancuri, grenade, precum și arme chimice. Primul Război Mondial se caracterizează ca fiind un război static datorită sistemului de tranșee implementat în toată Europa.

Războiul s-a încheiat în 1918 cu victoria Antantei și destrămarea a  patru mari imperii: Imperiul Otoman, Imperiul Austro-Ungar, Imperiul German și Imperiul Rus. În Rusia devastată de războiul civil și-a început ascensiunea noua ideologie a socialismului, luând naștere Uniunea Sovietică.
Multe schimbări teritoriale au avut loc, fiind create noi națiuni autonome în Europa și Orientul Mijlociu, iar alte colonii ale fostelor imperii înfrânte fiind transferate imperiilor și națiunilor învingătoare.
Pentru a preveni declanșarea unui alt război, națiunile învingătoare au pus bazele Ligii Națiunilor, dar care avea să se dovedească incapabilă și slabă.

Al Doilea Război Mondial (1939-1945)

Al Doilea Război Mondial a durat din 1939 până în 1945, fiind cel mai distructiv conflict militar din istorie și fiind singurul în care au fost utilizate arme atomice, fiind un adevărat război total de uzură.
Germania Nazistă a fost responsabilă pentru genociduri, cel mai notabil fiind Holocaustul, pentru uciderea a 6 milioane de evrei și al altor milioane de victime străine ucise în operațiuni militare. Germania este considerată  drept responsabilă  pentru declanșarea războiului în Europa datorită pretențiilor teritoriale ale lui Adolf Hitler conform conceptului fundamental al geopoliticii clasice germane "Lebensraum".  Germania  fusese anterior dezavantajată de condițiile dure impuse de puterile învingătoare, efectiv falimentand-o. Hitler, carismatic, naționalist și populist, a obținut încrederea și susținerea populației germane, dobândind puterea prin mijloace democratice și ajungând să instaureze un regim totalitar. A extins și dezvoltat Reich-ul german prin anexarea teritoriilor din Europa Centrală, Africa și Europa Vestică.
Între timp, în Uniunea Sovietică, pentru a-și consolida puterea și regimul totalitar, Iosif Stalin efectua epurări, genocide, execuții la scară largă și deportări, iar politicile sale brutale au dus la foamete  încât  zeci de milioane de sovietici au fost uciși. Inițial fiind într-o stare de înțelegere, cele două mari puteri au ajuns să se confrunte pentru supremația asupra estului Europei.
În același timp, Italia aspiră să refacă vechiul Imperiu Roman și a devenit părtaș la atrocitățile comise de germanii naziști.
În extremul orient, în urmă  agresivității de care a dat dovadă Imperiul Japonez în campaniile militare de expansiune și masacre,  America i-a tăiat aprovizionarea cu combustibili.  Japonezii, pentru a se răzbună, au comis atacul de la Pearl Harbour, ceea ce a determinat America  să iasă din neutralitate și să intervină, războiul, inițial european, devenind global.
Intervenția Americii și întărirea alianței acesteia dintre Marea Britanie și Uniunea Sovietică au dus la înfrângerea  puterilor Axei.

În urmă celui de-al doilea război mondial, cel de-al treilea Reich German a fost desființat, iar Germania divizată în două state separate, Imperiul Japonez și-a pierdut toate teritoriile în urmă confruntărilor crâncene și celor două bombardamente nucleare de la Hiroshima și Nagasaki efectuate de americani, multe state europene participante la război au fost eliberate de americani, cât și de sovietici de sub regimurile fasciste asociate cu Germania Nazistă, iar Franța și Imperiul Britanic și-au pierdut imperiile coloniale datorită distrugerilor, costurilor imense și emancipării mișcărilor pentru decolonizare.
Tabăra Aliaților, la conducerea președintelui Franklin Roosevelt, Winston Churchill și Iosif Stalin, a ieșit victorioasă din război și și-a permis să împartă sferele de influență în perioada postbelică.
Statele Unite ale Americii și-au asigurat poziția de superputere globală, alături de Uniunea Sovietică ce i-a devenit  superputerea rivală, ambele luptând în următoarele decenii pentru supremația globală.
În Orientul extrem, după anihilarea Imperiului Japonez, o nouă putere lua amploare, Republica Populară Chineză, care din păcate avea să comită mai multe agresiuni și masacre decât Imperiul Japonez. Asta în vreme ce Uniunea Sovietică a implementat conducerea socialistă în Europa de Est.

Instituții mondiale ca Organizația Națiunilor Unite au fost create pentru a colectiviza afacerile internaționale, cu scopul explicit de a preveni declanșarea unui alt război mondial. Războiul a comis multe distrugeri, dar a contribuit și la progresul tehnologic enorm ce vor avea efect profund în timp de pace, precum apariția vehiculelor aeriene de mare viteză, penicilină, energia nucleară și  calculatoarele electronice.

Al Treilea Război Mondial

Războiul Rece (1947-1991 )

Termenul de Al Treilea Război Mondial  este o noțiune care desemnează un eventual război global, care ar urmă să fie purtat la scară globală și care ar avea loc ca o continuare a celui de-al Doilea Război Mondial. Acest presupus război a fost anticipat și plănuit în perioada Războiului Rece de autorități civile și militare din mai multe țări și, de asemenea, este un subiect foarte des întâlnit în literatură de ficțiune. Se presupune că se vor folosi arme atomice care vor duce la distrugerea Pământului. În perioada conflictului dintre Statele Unite și Uniunea Sovietică, denumit generic Războiul Rece, se credea că este posibil ca al Treilea Război Mondial să izbucnească din cauza rivalităților dintre aceste două superputeri.
Datorită distrugerii reciproce asigurate, un al Treilea Război Mondial în care s-ar folosi arme nucleare ar duce la distrugeri colosale.

Deși a fost în mare o stare generală de tensiune dintre SUA și URSS,  omenirea a fost martoră la declanșarea multor războaie distrugătoare în mai multe regiuni ale planetei în care s-au implicat direct sau indirect ambele superputeri în confruntări proxy. Nu s-a ajuns niciodată la o confruntare directă, existând doar momente ce încingeau tensiunile dintre cele două superputeri, cum a fost criza rachetelor din Cuba disputata  dintre președintele american John F. Kennedy și liderul sovietic Nikita Hrusciov și rezolvată pașnic după negocieri.

Cu toate astea, SUA și URSS s-au confruntat în mod indirect prin intermediari în războaie ca Războiul din Coreea (1950–1953), Criza Berlinului din 1961, Criza Suezului (1962), Războiul din Vietnam (1959–1975), Războiul de Iom Kippur (1973), Războiul Afgano-Sovietic (1979–1989), doborârea cursei KAL 007 (1983) și exercițiul militar NATO "Able Archer" (1983)  și multe alte războaie civile din Africa de după decolonizare și colapsul imperiului colonial britanic și al celui francez.

Europa a fost împărțită în două părți, linia de marcație fiind cortina de fier, culminată cu Zidul Berlinului, Europa de Vest, aflându-se în sfera SUA, considerată drept blocul de vest, lumea întâi sau lumea liberă care era alcătuită din state dezvoltate,  și  Europa de Est aflată în sfera URSS, cunoscută ca blocul de est, lumea a două  sau gulagul socialist, alcătuită din state în curs de dezvoltare.
Lumea a treia era alcătuită din state subdezvoltate, state nealiniate  sau statele extra-europene, unele dintre ele fiind aliatele Americii (precum Japonia, Coreea de Sud, Australia, Canada, Turcia, Arabia Saudită, Mexic, Brazilia), altele fiind aliatele Uniunii Sovietice (China, Vietnam, Coreea de Nord, Cuba).
SUA a căutat să-și impună dominația prin extinderea pieței libere, globalizare, instaurarea regimurilor democratice  și capitalizare. URSS și-a impus dominația prin susținerea mișcărilor ideologice marxiste în multe state promițând progres și bunăstare  prin intervenționism și etatism în defavoarea nerespectării drepturilor fundamentale ale omului precum dreptul la proprietate și presă liberă.
Concurența a devenit acerbă în cursa înarmărilor  cât și în cursa spațială pentru primul pas pe Lună, competiție câștigată de SUA. Un rol important l-au jucat serviciile secrete, confruntându-se, în primul rând, cele americane (CIA, NSA) cu KGB-ul sovietic. Au fost implicate, însă, și serviciile secrete vest-europene (britanice, vest-germane, franceze, italiene, etc.) și est-europene (Securitatea, STASI, etc.). Denumirea de "Război Rece" mai provine și din faptul că a fost purtat între foștii aliați din războiul împotriva regimului totalitar nazist. Din punct de vedere al mijloacelor utilizate, Războiul Rece a fost un conflict în care s-au utilizat presiunea diplomatică, militară, economică, ajutorul pe scară largă pentru statele-client, manevrele diplomatice, spionajul, curse ale înarmărilor convenționale și nucleare, coaliții militare, rivalitate la evenimentele sportive, competiție tehnologică, campanii masive de propagandă, asasinatul, operațiunile militare de intensitate mică și iminența unui război pe scară mare.

La mijlocul anilor 1980, conducerea sovietică a lui Mihail Gorbaciov din URSS a introdus reformele de liberalizare, presat de cererile de independență națională a sateliților sovietici din estul Europei, cât și de faliment. Ronald Reagan a initiat Inițiativa Strategică de Apărare pentru crearea unui scut antirachetă nordamerican împotriva atacurilor cu rachete balistice intercontinentale lansate de un stat inamic, în special de Uniunea Sovietică, pe care o considera un imperiu al răului.

După lunga dictatura a lui Mao Zedong, care a comis atrocități prin Marele Salt Înainte sau Revoluția Culturală, China și-a început procesul de liberalizare sub Deng Xiaoping, menținând însă regimul comunist la conducere.
Războiul Rece s-a atenuat odată cu prăbușirea regimurilor comuniste din Europa Centrală și de Est , urmată, doi ani mai târziu, în decembrie 1991, și de destrămarea Uniunii Sovietice și crearea Federației Ruse  și dobândirea independenței a multor state. Statele central-europene și est-europene (care, timp de patru decenii, se aflaseră sub dominația sovieticǎ), s-au democratizat și au ales să se integreze în NATO și Uniunea Europeană. Lumea care a rămas este dominată de o singură superputere, SUA. Această situație este, de regulă, descrisă drept hegemonie globală a SUA într-o lume unipolară, deși unii autori consideră că lumea actuală este multipolară.

După douǎ decenii de destindere a relațiilor americano-ruse, urmatǎ, în 2008, de rǎcirea relațiilor diplomatice - consecință a rǎzboiului din Georgia - și, mai ales, pe fondul crizei din Ucraina, tensiunile, ostilitățile și rivalitǎțile anterioare dintre cele două puteri s-au acutizat, astfel că, în 2014, a reizbucnit Războiul Rece între Statele Unite ale Americii/ statele membre ale Uniunii Europene/ Organizației Tratatului Atlanticului de Nord și Federația Rusă  și aliații acesteia. Acest rǎzboi este cunoscut în Mass-Media ca "Războiul Rece 2.0". însă, spre deosebire de Războiul Rece anterior, de această dată, Germania reunificată are un rol geopolitic major în Europa.
În 2018, SUA a impus tarife produselor fabricate în China, declanșând un război comercial.

Războiul împotriva terorismului (2001-prezent)

Războiul împotriva terorismului, cunoscut și sub numele de Războiul global al terorismului, este o campanie militară asimetrică internațională lansată de guvernul Statelor Unite și administrația prezidențială a lui George W. Bush jr. după atacurile teroriste de la New York din 11 septembrie 2001 efectuate de rețeaua teroristă al-Qaeda, coordonate de Osama bin Laden.
SUA și coaliția națiunilor anti-terorism au condus operațiuni în Afganistan  distrugând taberele al-Qaeda și înlăturând regimul taliban, precum și în nord-vestul Pakistanului în căutarea autorului atentatelor, Osama Bin Laden, fiind căutat și ucis abia după 10 ani.
Între timp, sub acuzația că ar fi deținut arme chimice, regimul dictatorului Saddam Hussein a fost înlăturat de la putere din Irak în urma intervenției militare dusă de SUA și au fost desfășurate primele alegeri democratice.
În ciuda eforturilor de democratizare și menținere a păcii de către trupele americane, insurgențele din Irak cât și din Afganistan n-au făcut decât să prelungească războiul.  În 2011, Primăvară Arabă a fost declanșată în mai multe state din Orientul Apropiat și Mijlociu, regimuri dictatoriale fiind înlăturate. În Siria însă, regimul lui Bashar al-Assad a continuat să reziste rebelilor, finanțați de administrația americană prezidențială a lui Barack Obama. O nouă grupare teroristă și-a făcut apariția, Statul Islamic al Irakului și Levantului, anexand teritorii din Irak și Siria, proclamându-se califat.
Teroriști ISIS  au comis masacre și atrocități în Siria și Irak, ceea ce a determinat populațiile acestora să se refugieze în Europa. În marile orașe europene, mai ales la Paris, Londra, Berlin, Barcelona și Bruxelles  adepții grupării ISIS și islamici radicali au comis atentate sângeroase.
SUA și Rusia au desfășurat atacuri aeriene împotriva țintelor ISIS din Siria și Irak, Statul Islamic fiind anihilat, iar teritoriile Irakului și Siriei fiind eliberate.
SUA sub administrația președinților Obama și Trump au condus operațiuni militare pentru combaterea terorismului în Nigeria, Irak, Afganistan, Somalia, Pakistan, Libia, Egipt și Yemen, dar și în Siria împotriva regimului Assad când a utilizat arme chimice asupra populației civile din tabăra opozantă.

Războiul continuă și în prezent.

Statistici

Note

Vezi și
 Primul Război Mondial
 Al Doilea Război Mondial
 Al Treilea Război Mondial - un conflict ipotetic, în viitor

Război după tip

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A tip o' the hat to Jaime Ullinger of the BioAnthropology News group on facebook, and to Katha Pollitt of The Nation, for bringing this to my attention.
Each year the Gallup organization polls the US public on their beliefs with respect to evolution. The question, asked each year, offers three alternative 'takes' on reality.
If you add up the two alternatives with 'God' in the mix, there has been an almost imperceptible fall-off over the past 20 years. But essentially the viewpoint of the American public is unchanged in that time. The [for me] ultimate response is to alter only slightly the words of a bumper sticker I remember seeing a long time ago.
Beam me up, Scotty. There's less intelligent life down here than we thought.
Seriously? Between 44% and 47% of a cross-section of the US population are young-earth Christianists. I'm no math whiz, but that's a scary lot of WHACK-Os walking the streets. As Ms. Pollitt points out in her editorial, it's no wonder the right is able to pull the wool over the eyes of their supporters on so many issues linked to the clear rejection of science on the part of half the voters. The author of the Nation piece quotes a perspicacious politico with a lovely turn of mind.
So, get going all my subversive friends! Forget talking about global weirding and start talkin' 'bout the next flood! That'll get their attention!
I believe the the results are similar when Americans are asked if they believe in ghosts, UFOs, ESP, telekinesis, etc. Half to two-thirds do. So it's not just Christianism that scientists are up against.
I can fudge these numbers to find something hopeful. If you lump together all the responses that involve god, the trend line is clearly negative, but the r-squared value is only 0.2. If you lump together all the responses that include evolution the trendline is clearly positive and the r-squared is 0.33. By my reckoning the numbers will be about equal in 2117. Does the Maya calendar go beyond that?
I do not know how to copy the graph into this, but I will send it to Rob.
@Spawn. Sadly, you're right. And don't forget faeries, leprechauns, zombies and vampires. If I can paraphrase the old real estate mantra, the answer is Education, Education, Education. As long as the corporatists are running the place, education will always take a back seat to the almighty buck.
@ Iain: If you retrocast that regression, does 100% belief in God-forcing appear at 4004 BC? Another informative benchmark, perhaps.
I am afraid you are wide of the mark Spawn (if I may call you that). The date when no one thought that evolution was the driving force behind the presence of humans was 1548--the year before the publication of the Book of Common Prayer. It was, of course, almost as big an influence on the English language as the King James Bible.
For the date when everyone thought god was involved you get back to a much more interesting date. I swear I did not fiddle this in the slightest, but you are not going to believe that the date was so recent. It was 1859. I'm not sure that that has had much of an influence. Oh, hang on, isn't that the date of publication of "On the origin of species"? Say it isn't so. To quote a great Research Scholar from the Ronin Institute: Coincidence? I hardly think so.
Spawn E. hits on something very important. Those polls do tend to dredge up stats that are pretty eye-popping no matter what the question sets are. The answers most likely reflect who they're actually polling and where they conduct the survey more than they show a representation of American beliefs.
Nice choice of colours, Iain. Another coincidence? I think not. Rebublican blue and Democratic red. Positively poetry!


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O Castelo de Lochranza () é um castelo atualmente em ruínas localizado na Ilha de Arran, North Ayrshire, Escócia.

História
Foi construído provavelmente por um membro da família Stewart de Menteith, sendo ocupado em 1614 pelas tropas de Jaime VI e mais tarde nesse século pelas forças de Oliver Cromwell.

Encontra-se classificado na categoria "A" do "listed building" desde 14 de abril de 1971.

Estrutura
O castelo original é do final do século XIII ou inicio do século XIV com dois pisos e uma pequena torre. O castelo atualmente tem três pisos e um sótão.

Ver também
Lista de castelos da Escócia

Lochranza
Ruínas da Escócia


text:
\section{Introduction}
\label{intr}

The class of all semigroup varieties forms a lattice under the following
naturally defined operations: for varieties $\mathcal X$ and $\mathcal Y$,
their \emph{join} $\mathcal{X\vee Y}$ is the variety generated by the
set-theoretical union of $\mathcal X$ and $\mathcal Y$ (as classes of
semigroups), while their \emph{meet} $\mathcal{X\wedge Y}$ coincides with the
class-theoretical intersection of $\mathcal X$ and $\mathcal Y$. This lattice
has been intensively studied for about four decades. A systematic overview of
the material accumulated here is given in the recent survey \cite
{Shevrin-Vernikov-Volkov-09}.

It is a common knowledge that the lattice \textbf{SEM} of all semigroup
varieties is divided into two large sublattices with essentially different
properties: the coideal \textbf{OC} of all \emph{overcommutative} varieties
(that is, varieties containing the variety of all commutative semigroups) and
the ideal of all \emph{periodic} varieties (that is, varieties consisting of
periodic semigroups).

The global structure of the lattice \textbf{OC} has been revealed by Volkov
in~\cite{Volkov-94}. It is proved there that this lattice decomposes into a
subdirect product of its certain intervals and each of these intervals is
anti-isomorphic to the congruence lattice of a certain unary algebra of a
special type (namely, of a so-called $G$-set; a basic information about
$G$-sets see in~\cite{McKenzie-McNulty-Taylor-87}, for instance). The exact
formulation of this result may be found also in~\cite[Theorem~5.1]
{Shevrin-Vernikov-Volkov-09}. We do not reproduce this formulation here
because we do not use it below.

There are several articles where special elements of different types in the
lattice \textbf{SEM} have been examined (see~\cite{Jezek-McKenzie-93,
Vernikov-07-cmod,Vernikov-07-lmod,Vernikov-08-lmod,Vernikov-08-umod1,
Vernikov-08-umod2,Vernikov-codistr,Vernikov-Shaprynskii-distr,
Vernikov-Volkov-06,Volkov-05}). We refer an interested reader to~\cite
[Section~14]{Shevrin-Vernikov-Volkov-09} for an overview of the most part of
results obtained in these articles.

Recall that an element $x$ of a lattice $\langle L;\ \vee,\wedge\rangle$ is
called \emph{distributive} if
$$\forall y,z\in L\colon\quad x\vee(y\wedge z)=(x\vee y)\wedge(x\vee z);$$
\emph{standard} if
$$\forall y,z\in L\colon\quad(x\vee y)\wedge z=(x\wedge z)\vee(y\wedge z);$$
\emph{neutral} if, for all $y,z\in L$, the sublattice of $L$ generated by
$x$, $y$, and $z$ is distributive. \emph{Codistributive} [\emph
{costandard}\,] elements are defined dually to distributive [respectively
standard] ones. An extensive information about elements of all these five
types in abstract lattices may be found in~\cite[Section~III.2]{Gratzer-98},
for instance. Note that any [co]standard element is [co]distributive, and an
element is neutral if and only if it is standard and costandard
simultaneously (see~\cite [Theorem~III.2.5]{Gratzer-98}, for instance). On
the other hand, a [co]distributive element may be not [co]standard, while a
[co]standard element may be not neutral.

A complete description of neutral elements in the lattice \textbf{SEM} has
been given in~\cite[Proposition~4.1]{Volkov-05} (see also~\cite[Theorem~14.2]
{Shevrin-Vernikov-Volkov-09}). In~\cite{Vernikov-Shaprynskii-distr}, all
distributive elements in \textbf{SEM} are completely determined. In~\cite
{Vernikov-codistr}, quite a strong necessary condition for semigroup
varieties to be a codistributive element in \textbf{SEM} is obtained. In
particular, all varieties with each of these three properties (except the
trivial extreme case of the variety $\mathcal{SEM}$ of all semigroups) turn
out to be periodic varieties.

So, an examination of special elements of all the mentioned types in the
lattice \textbf{SEM} gives no any information concerning the lattice \textbf
{OC}. Aiming to obtain some new knowledge about this lattice, it is natural
to investigate its special elements.

Such investigations have been started by the second author in~\cite
{Vernikov-01}. Five types of special elements (namely, distributive,
codistributive, standard, costandard, and neutral elements) in the lattice
\textbf{OC} have been considered there. Unfortunately, it turns out that
considerations in~\cite{Vernikov-01} contain a gap, and the main result of
this article is incorrect. Namely, it was proved in~\cite{Vernikov-01} that,
for an overcommutative semigroup variety, the properties of being a
distributive element of \textbf {OC}, of being a codistributive element of
\textbf {OC}, of being a standard element of \textbf {OC}, of being a
costandard element of \textbf{OC}, and of being a neutral element of \textbf
{OC} are equivalent. This result of~\cite{Vernikov-01} is true. But, besides
that, the main result of~\cite{Vernikov-01} contains a list of all
overcommutative varieties that possess the five mentioned properties.
Unfortunately, this list turns out to be non-complete. All varieties from the
list really have all the mentioned properties, but there are many other such
varieties. The objective of this article is to give a correct description of
distributive, codistributive, standard, costandard, and neutral elements in
the lattice \textbf{OC}.

The article is structured as follows. In Section~\ref{prel}, we introduce a
necessary notation and formulate the main result of the article (Theorem~\ref
{main}). In Section~\ref{order}, we prove several auxiliary facts.
Sections~\ref{nec} and~\ref{suff} are devoted to the proof of Theorem~\ref
{main}. In Section~\ref{optim}, we show that this theorem can not be
improved, in a sense. Finally, in Section~\ref{quest}, we formulate some open
problems.

\section{Preliminaries and summary}
\label{prel}

We denote by $F$ the free semigroup over a countably infinite alphabet
$\{x_1,x_2$, $\dots,x_n,\dots\}$. As usual, elements of $F$ are called \emph
{words}. By $F^1$ we denote the semigroup $F$ with the empty word ajoined.
The symbol $\equiv$ stands for the equality relation on $F$ and $F^1$. If $u$
is a word, then $\ell(u)$ denotes the length of $u$, $\ell_i(u)$ is the
number of occurrences of the letter $x_i$ in $u$, $c(u)$ stands for the set
of all letters occurring in $u$, and $n(u)=|c (u)|$ is the number of letters
occurring in $u$. An identity $u\approx v$ is called \emph{balanced} if
$\ell_i(u)=\ell_i(v)$ for all $i$. It is a common knowledge that if an
overcommutative variety satisfies some identity then this identity is
balanced.

Let $m$ and $n$ be integers with $2\le m\le n$. A \emph{partition of the
number $n$ into $m$ parts} is a sequence of positive integers $(\ell_1,
\ell_2,\dots,\ell_m)$ such that
$$\ell_1\ge\ell_2\ge\cdots\ge\ell_m\quad\text{and}\quad\sum_{i=1}^m\ell_i=n
\ldotp$$
The numbers $\ell_1,\ell_2,\dots,\ell_m$ are called \emph{components} of the
partition $\lambda$. We denote by $\Lambda_{n,m}$ the set of all partitions
of the number $n$ into $m$ parts and by $\Lambda$ the union of the sets
$\Lambda_{n,m}$ for all natural numbers $m$ and $n$ with $2\le m\le n$. If
$\lambda\in\Lambda_{n,m}$ then we denote the numbers $n$ and $m$ by $n
(\lambda)$ and $m(\lambda)$ respectively.

If $u$ is a word then we denote by $\partition(u)$ the partition of the
number $\ell(u)$ into $n(u)$ parts consisting of integers $\ell_i(u)$ for all
$i$ such that $x_i\in c(u)$ (the numbers $\ell_i(u)$ are placed in
$\partition(u)$ in non-increasing order). If $u\approx v$ is a balanced
identity then, obviously, $\ell(u)=\ell(v)$, $n(u)=n(v)$, and $\partition(u)=
\partition(v)$. We call the partition $\partition(u)$ a \emph{partition} of
the identity $u\approx v$. We denote the numbers $\ell(u)=\ell(v)$ and $n(u)=
n(v)$ by $\ell(u\approx v)$ and $n(u\approx v)$ respectively, and the
partition $\partition(u)=\partition(v)$ by $\partition(u\approx v)$.

Let $\lambda=(\ell_1,\ell_2,\dots,\ell_m)\in\Lambda_{n,m}$. We denote by
$W_{n,m,\lambda}$, or simply $W_\lambda$, the set of all words $u$ such that
$\ell(u)=n$, $c(u)=\{x_1,x_2,\dots,x_m\}$, $\ell_i(u)\ge\ell_{i+1}(u)$ for
all $i=1,2,\dots,m-1$, and $\partition(u)=\lambda$. It is evident that every
balanced identity $u\approx v$ with $\ell(u\approx v)=n$, $n(u\approx v)=m$,
and $\partition(u\approx v)=\lambda$ is equivalent to some identity $s\approx
t$ where $s,t\in W_{n,m,\lambda}$.

We call sets of the kind $W_{n,m,\lambda}$ \emph{transversals}. We say that
an overcommutative variety $\mathcal V$ \emph{reduces} [\emph{collapses}] a
transversal $W_{n,m,\lambda}$ if $\mathcal V$ satisfies some non-trivial
identity [all identities] of the kind $u\approx v$ with $u,v\in
W_{n,m,\lambda}$. An overcommutative variety $\mathcal V$ is said to be \emph
{greedy} if it collapses any transversal it reduces. The following assertion
has been proved in~\cite{Vernikov-01}.

\begin{proposition}
\label{greedy}
An overcommutative semigroup variety is a distributive \textup
[co\-distributive, standard, costandard, neutral\textup] element of the
lattice $\mathbf{OC}$ if and only if it is greedy.\qed
\end{proposition}

This assertion was not formulated in~\cite{Vernikov-01} explicitly but it
directly follows from the proof of Theorem~2 in~\cite{Vernikov-01} (and
the corresponding part of the proof in~\cite{Vernikov-01} is correct).

It is an appropriate place here to indicate the error made in~\cite
{Vernikov-01}. Let $m$ and $n$ be positive integers with $2\le m\le n$ and
$\lambda\in\Lambda_{n,m}$. A semigroup variety given by an identity system
$\Sigma$ is denoted by $\var\Sigma$. We put
\begin{align*}
&\mathcal X_n=\var\,\{u\approx v\mid\ \text{the identity}\ u\approx v\ \text
{is balanced and}\ \ell(u\approx v)\ge n\},\\
&\mathcal X_{n,m}=\mathcal X_{n+1}\wedge\var\,\{u\approx v\mid\ \text{the
identity}\ u\approx v\ \text{is balanced},\ \ell(u\approx v)=n,\\
&\phantom{\mathcal X_{n,m}=\mathcal X_{n+1}\wedge\var\,\{u\approx v\mid\ \ }
\text{and}\ n(u\approx v)\le m\},\\
&\mathcal X_{n,1}=\mathcal X_{n+1},\\
&\mathcal X_{n,m,\lambda}=\mathcal X_{n,m-1}\wedge\var\{u\approx v\mid u,v\in
W_{n,m,\lambda}\}\ldotp
\end{align*}
It is claimed in~\cite{Vernikov-01} without any proof that an overcommutative
variety is greedy if and only if it coincides with one of the varieties
$\mathcal{SEM}$, $\mathcal X_n$, $\mathcal X_{n,m}$ or $\mathcal
X_{n,m,\lambda}$. Combining this claim with Proposition~\ref{greedy}, we
obtain the main result of~\cite{Vernikov-01}: the varieties $\mathcal{SEM}$,
$\mathcal X_n$, $\mathcal X_{n,m}$, $\mathcal X_{n,m,\lambda}$, and only they
are [co]distributive, [co]standard, and neutral elements in \textbf{OC}. In
actual fact, it is true that all these varieties are elements of the
mentioned types in \textbf{OC}. But the list of [co]distributive,
[co]standard, and neutral elements in \textbf{OC} is not exhausted by the
varieties $\mathcal{SEM}$, $\mathcal X_n$, $\mathcal X_{n,m}$, and $\mathcal
X_{n,m,\lambda}$. There are many other varieties with such a property.
Exactly this fact has been so unfortunately overseen in~\cite{Vernikov-01}.

For a partition $\lambda=(\ell_1,\ell_2,\dots,\ell_m)\in\Lambda_{n,m}$, we
define numbers $q(\lambda)$, $r(\lambda)$, and $s(\lambda)$ by the following
way:
\begin{align*}
&q(\lambda)\ \text{is the number of}\ \ell_i\text{'s with}\ \ell_i=1\ \text
{(if}\ \ell_m>1\ \text{then}\ q(\lambda)=0\text{);}\\
&r(\lambda)\ \text{is the sum of all}\ \ell_i\text{'s with}\ \ell_i>1\ \text
{(if}\ \ell_1=1\ \text{then}\ r(\lambda)=0\text{);}\\
&s(\lambda)=\max\,\{r(\lambda)-q(\lambda)-\delta,0\}
\end{align*}
where
$$\delta=
\begin{cases}
0&\text{whenever}\ n=3,m=2,\ \text{and}\ \lambda=(2,1),\\
1&\text{otherwise}\ldotp
\end{cases}$$
If $k$ is a non-negative integer then $\lambda^k$ stands for the following
partition of $n+k$ into $m+k$ parts:
$$\lambda^k=(\ell_1,\ell_2,\dots,\ell_m,\underbrace{1,\dots,
1}_{k\ \text{times}})$$
(in particular, $\lambda^0=\lambda$).

For a partition $\lambda\in\Lambda_{n,m}$, we put
$$\mathcal W_{n,m,\lambda}=\var\,\{\,u\approx v\mid u,v\in W_{n,m,\lambda}\}
\quad\text{and}\quad\mathcal S_\lambda=\bigwedge_{i=0}^{s(\lambda)}\mathcal
W_{n+i,m+i,\lambda^i}\ldotp$$
Sometimes we will write $\mathcal W_\lambda$ rather than $\mathcal
W_{n,m,\lambda}$.

The main result of the article is the following

\begin{theorem}
\label{main}
For an overcommutative semigroup variety $\mathcal V$, the following are
equivalent:
\begin{itemize}
\item[(i)]$\mathcal V$ is a distributive element of the lattice $\mathbf
{OC}$;
\item[(ii)]$\mathcal V$ is a codistributive element of the lattice $\mathbf
{OC}$;
\item[(iii)]$\mathcal V$ is a standard element of the lattice $\mathbf{OC}$;
\item[(iv)]$\mathcal V$ is a costandard element of the lattice $\mathbf{OC}$;
\item[(v)]$\mathcal V$ is a neutral element of the lattice $\mathbf{OC}$;
\item[(vi)]either $\mathcal{V=SEM}$ or $\mathcal V=\bigwedge\limits_{i=1}^k
\mathcal S_{\lambda_i}$ for some partitions $\lambda_1,\lambda_2,\dots,
\lambda_k\in\Lambda$.
\end{itemize}
\end{theorem}

The following claim was formulated in \cite{Vernikov-01} as a corollary of
the main result of that article. Theorem \ref{main} shows that the claim is
correct.

\begin{corollary}
\label{countable}
The set of all \textup[co\textup]distributive elements of the lattice
$\mathbf{OC}$ is countably infinite.\qed
\end{corollary}

This corollary is of some interest because the set of all overcommutative
semigroup varieties is well known to be uncountably infinite. On the other
hand, it is interesting to note that the set of all neutral elements in the
lattice \textbf{OC} is infinite, while the set of all neutral elements in the
lattice \textbf{SEM} consists of 5 varieties only~\cite[Proposition~4.1]
{Volkov-05}.

In view of Proposition~\ref{greedy}, Theorem~\ref{main} is equivalent to the
following

\begin{proposition}
\label{reduce}
An overcommutative semigroup variety $\mathcal V$ satisfies the
condition~\textup{(vi)} of Theorem~\textup{\ref{main}} if and only if it is
greedy.
\end{proposition}

It is this claim that will be verified in Sections~\ref{nec} and~\ref{suff}
(in fact, we prove the `only if' and `if' parts of Proposition~\ref{reduce}
in Sections~\ref{nec} and~\ref{suff} respectively). To prepare this proof, we
introduce some order relation on the set $\Lambda$ and consider some
properties of this relation in Section~\ref{order}.

\section{An order relation on the set $\Lambda$}
\label{order}

Let $\lambda=(\ell_1,\ell_2,\dots,\ell_m)\in\Lambda_{n,m}$ where $m\ge3$ and
$1\le i<j\le m$. We denote by $U_{i,j}(\lambda)$ the partition of the number
$n$ into $m-1$ parts with the components $\ell_1$, $\ell_2$, \dots,
$\ell_{i-1}$, $\ell_{i+1}$, \dots, $\ell_{j-1}$, $\ell_{j+1}$, \dots,
$\ell_m$, and $\ell_i+\ell_j$ (these components are written in $U_{i,j}
(\lambda)$ in non-increasing order). We will say that the partition $U_{i,j}
(\lambda)$ is obtained from $\lambda$ by the \emph{union of components}
$\ell_i$ and $\ell_j$. The partitioin obtained from $\lambda$ by a finite
(may be empty) set $S$ of unions of components is denoted by $U_S(\lambda)$;
in particular, $U_\varnothing(\lambda)=\lambda$.

We introduce a binary relation $\preceq$ on the set $\Lambda$ by the
following rule:
$$\lambda\preceq\mu\ \text{if and only if}\ \mu=U_S\bigl(\lambda^k\bigr)\
\text{for some}\ S\ \text{and}\ k\ldotp$$
The principal property of the relation $\preceq$ is given by the following

\begin{lemma}
\label{le is order}
The relation $\preceq$ is a partial order on the set $\Lambda$.
\end{lemma}

\begin{proof}
Reflexivity of $\preceq$ is evident because $\lambda=U_\varnothing\bigl
(\lambda^0\bigr)$. The claim that $\preceq$ is transitive also is evident
because if $\mu=U_S\bigl(\lambda^k\bigr)$ and $\nu=U_T\bigl(\mu^\ell\bigr)$
then $\nu=U_{S\cup T}\bigl(\lambda^{k+\ell}\bigr)$. To prove that $\preceq$
is antisymmetric, we suppose that $\lambda\preceq\mu$ and $\mu\preceq\lambda$
for some $\lambda,\mu\in\Lambda$. Then $\mu=U_S\bigl(\lambda^k\bigr)$ and
$\lambda=U_T\bigl(\mu^\ell\bigr)$ for some $S$, $T$, $k$, and $\ell$. Let
$n(\lambda)=n$ and $n(\mu)=q$. Then $q=n+k$ and $n=q+\ell$. Therefore, $q=q+
k+\ell$, whence $k+\ell=0$. This means that $k=\ell=0$. Thus, $\mu=U_S
(\lambda)$ and $\lambda=U_T(\mu)$. If $S\ne\varnothing$ then $r<m$. But $m\le
r$ because $\lambda$ is obtained from $\mu$ by unions of components.
Therefore, $S=\varnothing$ and $\mu=U_\varnothing(\lambda)=\lambda$.
\end{proof}

Now we are going to show that the partial order $\preceq$ has some nice
properties that will be played the crucial role in Section~\ref{suff}. The
first such property is given by the following

\begin{lemma}
\label{min condition}
The partially ordered set $\langle\Lambda;\preceq\rangle$ satisfies the
descending chain condition.
\end{lemma}

\begin{proof}
Let $\lambda,\mu\in\Lambda$ and $\lambda\preceq\mu$. Put $n(\lambda)=n$ and
$n(\mu)=q$. Then $q\le n$. Evidently, the set
$$\bigcup_{\substack{q\le n\\
2\le r\le q}}\Lambda_{q,r}$$
is finite. Thus, there exists finitely many partitions $\mu$ with $\mu\preceq
\lambda$ only. This immediately implies the desirable conclusion.
\end{proof}

We define one more binary relation $\unlhd$ on the set $\Lambda$ by the
following rule. Let $\lambda,\nu\in\Lambda$, $\lambda=(\ell_1,\ell_2,\dots,
\ell_m)$, and $\nu=(n_1,n_2,\dots,n_k)$. Then $\lambda\unlhd\nu$ if and only
if $m\le k$ and $\ell_i\le n_i$ for all $i=1,2,\dots,m$. It is evident that
$\unlhd$ is a partial order on $\Lambda$. The following claim shows a
relationship between orders $\preceq$ and $\unlhd$.

\begin{lemma}
\label{2 orders}
Let $\lambda,\nu\in\Lambda$. If $\lambda\unlhd\nu$ then $\lambda\preceq\nu$.
\end{lemma}

\begin{proof}
Let $\lambda=(\ell_1,\ell_2,\dots,\ell_m)$ and $\nu=(n_1,n_2,\dots,n_k)$.
Then $m\le k$ and $\ell_i\le n_i$ for all $i=1,2,\dots,m$. Put $s=n(\nu)-n
(\lambda)$. It is evident that $s\ge0$. If $s=0$ then $\lambda=\nu$ and we
are done. Let now $s>0$. By the trivial induction, it suffices to consider
the case $s=1$. Then either $k=m+1$, $\ell_i=n_i$ for all $i=1,2,\dots,m$,
and $n_k=1$ or $k=m$, $n_i=\ell_i+1$ for some $i\in\{1,2,\dots,m\}$ and $n_j=
\ell_j$ for all $j\ne i$. It is evident that $\nu=U_\varnothing\bigl
(\lambda^1\bigr)$ in the former case, while $\nu=U_{i,m+1}\bigl(\lambda^1
\bigr)$ in the latter one. Thus, $\lambda\preceq\nu$ in any case.
\end{proof}

The second important property of the relation $\preceq$ is given by the
following

\begin{lemma}
\label{antichains}
The partially ordered set $\langle\Lambda;\preceq\rangle$ does not contain
infinite anti-chains.
\end{lemma}

\begin{proof}
Arguing by contradiction, suppose that $\Lambda$ contains an infinite
anti-chain $A_0$. Put $m_1=\min\,\{m(\lambda)\mid\lambda\in A_0\}$. Let us
fix a partition $\lambda_1=(\ell_1^1,\ell_2^1,\dots,\ell_{m_1}^1)\in A_0$. If
$\nu=(n_1,n_2,\dots,n_k)$ is an arbitrary partition from $A_0$ then
$\lambda_1\npreceq\nu$, whence $\lambda_1\ntrianglelefteq\nu$ by Lemma~\ref
{2 orders}. Since $m_1\le k$, this means that $\ell_i^1>n_i$ for some $i\in
\{1,2,\dots,m_1\}$. The set $A_0$ is infinite, while the index $i$ runs over
the finite set $\{1,2,\dots,m_1\}$. Hence there is an index $i_1\le m_1$ such
that $n_{i_1}<\ell_{i_1}^1$ for an infinite set of partitions $A_1\subseteq
A_0$. Put $j_1=\ell_{i_1}^1$.

Put $m_2=\min\,\{m(\lambda)\mid\lambda\in A_1\}$. Let us fix a partition
$\lambda_2=(\ell_1^2,\ell_2^2,\dots,\ell_{m_2}^2)\in A_1$. The same arguments
as in the previous paragraph show that there is a number $i_2\le m_2$ and an
infinite set $A_2\subseteq A_1$  such that $n_{i_2}<\ell_{i_2}^2$ for every
$\nu=(n_1,n_2,\dots,n_k)\in A_2$. Put $j_2=\ell_{i_2}^2$.

Continuing this process, we construct a sequence of infinite sets of
partitions $A_0\supseteq A_1\supseteq A_2\supseteq\cdots$, a sequence of
partitions $\{\lambda_s=(\ell_1^s,\ell_2^s,\dots,\ell_{m_s}^s)\mid s\in
\mathbb N\}$, and two sequences of numbers $\{i_s\mid s\in\mathbb N\}$ and
$\{j_s\mid s\in\mathbb N\}$ such that, for any $s\in\mathbb N$, the following
holds: $\lambda_s\in A_{s-1}$, $i_s\le m_s$, $j_s=\ell_{i_s}^s$, and
$n_{i_s}<j_s$ for any $\nu=(n_1,n_2,\dots,n_k)\in A_s$. The choice of the
partitions $\lambda_1,\lambda_2,\dots$ guarantees that if $p>q$ then
$$\ell_{i_q}^p<\ell_{i_q}^q=j_q\ldotp$$
In particular, if $p>q$ and $i_p=i_q$ then
$$j_p=\ell_{i_p}^p=\ell_{i_q}^p<j_q\ldotp$$
This means that all pairs of the kind $(i_s,j_s)$ are different. Furthermore,
if $i_p\ge i_q$ then $\ell_{i_p}^p\le\ell_{i_q}^p$ because all partitions
$\lambda_1,\lambda_2,\dots$ are non-increasing sequences of numbers.
Therefore, if $p>q$ and $i_p\ge i_q$ then
$$j_p=\ell_{i_p}^p\le\ell_{i_q}^p<j_q\ldotp$$
Put $i_r=\min\,\{i_s\mid s\in\mathbb N\}$, $j_t=\min\,\{j_s\mid s\in\mathbb
N\}$, and $h=\max\,\{r,t\}$. If $s>h$ then $i_s\ge i_r$ and $j_s\ge j_t$,
whence $j_s<j_r$ and $i_s<i_t$. We see that both the sequences $\{i_s\mid s
\in\mathbb N\}$ and $\{j_s\mid s\in\mathbb N\}$ are bounded. But this is
impossible because all pairs of the kind $(i_s,j_s)$ are different. The
contradiction completes the proof.
\end{proof}

\section{Proof of Proposition \ref{reduce}: necessity}
\label{nec}

Here we aims to verify that if an overcommutative variety satisfies the
condition~(vi) of Theorem~\ref{main} then it is greedy. We start with some
new notation and several auxiliary facts.

For arbitrary words $w_1,w_2$ and an identity system $\Sigma$, we write $w_1
\stackrel{\Sigma}{\longrightarrow}w_2$ if there exist $a,b\in F^1$, $s,t\in
F$, and an endomorphism $\zeta$ on $F$ such that $w_1\equiv a\zeta(s)b$, $w_2
\equiv a\zeta(t)b$, and the identity $s\approx t$ belongs to $\Sigma$. It is
a common knowledge that an identity $u\approx v$ follows from a system
$\Sigma$ if and only if there exists a sequence of words $w_0,w_1,\dots,
w_\ell$ such that
\begin{equation}
\label{deduct}
u\equiv w_0\stackrel{\Sigma}{\longrightarrow}w_1\stackrel{\Sigma}
{\longrightarrow}\cdots\stackrel{\Sigma}{\longrightarrow}w_\ell\equiv v\ldotp
\end{equation}
This sequence is called a \emph{deduction} of the identity $u\approx v$ from
$\Sigma$. Note that if $\Sigma$ consists of balanced identities then $\ell
(w_i)=\ell(u\approx v)$, $n(w_i)=n(u\approx v)$, and $\partition(w_i)=
\partition(u\approx v)$ for all $i=0,1,\dots,\ell$.

\begin{lemma}
\label{poor xi}
Let $u$ be a word and $\xi$ an endomorphism on $F$ such that $\ell(\xi(u))=
\ell(u)$. Then $\partition(u)\preceq\partition(\xi(u))$.
\end{lemma}

\begin{proof}
Put $\lambda=\partition(u)$. It is clear that $\xi(x)$ is a letter for every
letter $x$. The requirement conclusion follows from the following evident
observation: $\partition(\xi(u))=U_S\bigl(\lambda^0\bigr)$ where $S$ is a
finite (may be empty) set of unions of components of $\lambda$ corresponding
to letters from $c(u)$ with the same image under $\xi$.
\end{proof}

\begin{lemma}
\label{lambda less u=v}
Let $\lambda=(\ell_1,\ell_2,\dots,\ell_m)\in\Lambda_{n,m}$. If a non-trivial
identity $u\approx v$ holds in the variety $S_\lambda$ then $\lambda\preceq
\partition(u\approx v)$.
\end{lemma}

\begin{proof}
We put
$$\Sigma=\{f\approx g\mid\ \text{there is}\ i\in\{0,1,\dots,s(\lambda)\}\
\text{with}\ f,g\in W_{n+i,m+i,\lambda^i}\}\ldotp$$
Thus, $S_\lambda=\var\Sigma$. Let~\eqref{deduct} be a deduction of the
identity $u\approx v$ from $\Sigma$. Note that $\ell\ge1$ because the
identity $u\approx v$ is non-trivial. We have $w_0\equiv a\zeta(s)b$ and $w_1
\equiv a\zeta(t)b$ for some homomorphism $\zeta$ on $F$, some $a,b\in F^1$,
and some $s,t\in W_{n+i,m+i,\lambda^i}$ where $i\in\{0,1,\dots,s(\lambda)\}$.

If $\ell(a\zeta(s)b)=\ell(s)$ then the words $a$ and $b$ are empty and
$\ell(\zeta(s))=\ell(s)$. Here we may apply Lemma~\ref{poor xi} and conclude
that
\begin{align*}
\lambda\preceq U_\varnothing\bigl(\lambda^i\bigr)&=\lambda^i=\partition(s)
\preceq\partition(\zeta(s))=\\
&=\partition(a\zeta(s)b)=\partition(w_0)=\partition(u\approx v)\ldotp
\end{align*}
Therefore, $\lambda\preceq\partition(u\approx v)$, and we are done.

Suppose now that $\ell(a\zeta(s)b)>\ell(s)$. For each $j=1,2,\dots,m+i$, we
denote by $y_j$ the first letter of the word $\zeta(x_j)$. Thus, $\zeta(x_j)
\equiv y_ju_j$ for some $u_j\in F^1$. We have
\begin{align*}
\partition(a\zeta(s)b)&=\partition(a(\zeta(x_1))^{\ell_1}\cdots(\zeta
(x_m))^{\ell_m}\zeta(x_{m+1})\cdots\zeta(x_{m+i})b)=\\
&=\partition(a(y_1u_1)^{\ell_1}\cdots(y_mu_m)^{\ell_m}y_{m+1}u_{m+1}\cdots
y_{m+i}u_{m+i}b)=\\
&=\partition(y_1^{\ell_1}\cdots y_m^{\ell_m}y_{m+1}\cdots y_{m+i}u_1^{\ell_1}
\cdots u_m^{\ell_m}u_{m+1}\cdots u_{m+i}ab)\ldotp
\end{align*}
Let $k=\ell(a\zeta(s)b)-\ell(s)$. Put
\begin{align*}
&c_1\equiv y_1^{\ell_1}\cdots y_m^{\ell_m}y_{m+1}\cdots y_{m+i};\\
&c_2\equiv u_1^{\ell_1}u_2^{\ell_2}\cdots u_m^{\ell_m}u_{m+1}\cdots u_{m+i}a
b;\\
&c\equiv c_1c_2\equiv y_1^{\ell_1}\cdots y_m^{\ell_m}y_{m+1}\cdots y_{m+i}
u_1^{\ell_1}\cdots u_m^{\ell_m}u_{m+1}\cdots u_{m+i}ab;\\
&d\equiv x_1^{\ell_1}\cdots x_m^{\ell_m}x_{m+1}\cdots x_{m+i+k}\ldotp
\end{align*}
Since $\partition(c)=\partition(a\zeta(s)b)$ and $\partition(c_1)=\partition
(s)$, we have $\ell(c)=\ell(a\zeta(s)b)$ and $\ell(c_1)=\ell(s)$. Besides
that, $\ell(c)=\ell(c_1)+\ell(c_2)$. Therefore,
$$\ell(c_2)=\ell(c)-\ell(c_1)=\ell(a\zeta(s)b)-\ell(s)=k\ldotp$$
It is convenient for us to rewrite the word $c_2$ in the form $c_2\equiv z_1
z_2\dots z_k$ where $z_1,z_2,\dots,z_k$ are (not necessarily different)
letters. Let $\xi$ be an endomorphism on $F$ such that
$$\xi(x_j)\equiv
\begin{cases}
y_j&\text{whenever}\ 1\le j\le m+i,\\
z_{j-m-i}&\text{whenever}\ m+i+1\le j\le m+i+k\ldotp
\end{cases}$$
Then $c\equiv\xi(d)$. It is clear that
$$\ell(d)=\ell(c_1)+k=\ell(s)+\ell(a\zeta(s)b)-\ell(s)=\ell(a\zeta(s)b)=\ell
(c)=\ell(\xi(d))\ldotp$$
Now we may apply Lemma~\ref{poor xi} and conclude that
\begin{align*}
\lambda\preceq U_\varnothing\bigl(\lambda^k\bigr)&=\lambda^k=\partition(d)
\preceq\partition(\xi(d))=\partition(c)=\\
&=\partition(a\zeta(s)b)=\partition(w_0)=\partition(u\approx v)
\ldotp
\end{align*}
Therefore, $\lambda\preceq\partition(u\approx v)$, and we are done.
\end{proof}

\begin{lemma}
\label{W_lambda in W_mu}
Let $\lambda,\mu\in\Lambda$. If $\mu=U_S(\lambda)$ for some finite set $S$
of unions of components then $\mathcal W_\lambda\subseteq\mathcal W_\mu$.
\end{lemma}

\begin{proof}
Let $\lambda=(\ell_1,\ell_2,\dots,\ell_m)\in\Lambda_{n,m}$. By the trivial
induction, it suffices to consider the case when $\mu=U_{i,j}(\lambda)$ for
some $i$ and $j$. We have to verify that if $u,v\in W_{n,m-1,\mu}$ then the
identity $u\approx v$ holds in $\mathcal W_\lambda$. Since $\partition(u
\approx v)=U_{i,j}(\lambda)$, there is a letter $x_k$ with $\ell_k(u)=\ell_k
(v)=\ell_i+\ell_j$. Let $x_p$ and $x_q$ be some letters such that $x_p,x_q
\notin c(u)$. One can change the first $\ell_i$ occurences of $x_k$ in $u$
and in $v$ by $x_p$, while the last $\ell_j$ occurences of $x_k$ in $u$ and
in $v$ by $x_q$. We obtain some identity $s\approx t$ with $\partition(s
\approx t)=\lambda$. Hence the variety $\mathcal W_{n,m,\lambda}$ satisfies
$s\approx t$. If we substitute $x_k$ for $x_p$ and $x_q$ in $s\approx t$ then
we return to the identity $u\approx v$. Therefore, $u\approx v$ follows from
$s\approx t$, whence $u\approx v$ holds in $\mathcal W_{n,m,\lambda}$.
\end{proof}

Recall that a letter $x_i$ is called \emph{simple in a word} $u$ if $\ell_i
(u)=1$.

\begin{lemma}
\label{s(lambda)=0}
If $\lambda\in\Lambda_{n,m}$ and $s(\lambda)=0$ then $\mathcal
W_{n,m,\lambda}\subseteq\mathcal W_{n+k,m+k,\lambda^k}$ for any positive
integer $k$.
\end{lemma}

\begin{proof}
The definition of the number $s(\lambda)$ immediately implies that if $s
(\lambda)=0$ then $s\bigl(\lambda^k\bigr)=0$ as well for any $k>0$. This
observation implies that, by the trivial induction, it suffices to consider
the case $k=1$.

First of all, we note that $\lambda\ne(2,1)$ because $s(\lambda)=1$
otherwise. For brevity, put $q=q(\lambda)$, $r=r(\lambda)$, $s=s(\lambda)$,
and $t=m-q$. Since $\lambda\ne(2,1)$, we have $\delta=1$ and therefore, $s=
\max\,\{r-q-1,0\}$. The equality $s=0$ implies now that $r-q-1\le0$, that is
\begin{equation}
\label{r<q+2}
r\le q+1\ldotp
\end{equation}
Suppose that $q\le1$. Then $r\le2$. Let $\lambda=(\ell_1.\ell_2,\dots,
\ell_m)$. The definition of the number $r(\lambda)$ and the inequality $r\le
2$ imply that either $\ell_1=\ell_2=\cdots=\ell_m=1$ or $\ell_1=2$ and
$\ell_2=\cdots=\ell_m=1$. Since $\lambda\ne(2,1)$, this implies that $q\ge2$.
We have a contradiction with the inequality $q\le1$. Therefore, $q\ge2$. This
implies that every word from the transversal $W_{n+1,m+1,\lambda^1}$ has at
least three simple letters.

We need to verify that any identity of the kind $u\approx v$ with $u,v\in
W_{n+1,m+1,\lambda^1}$ holds in $W_{n,m,\lambda}$. It suffices to check that
if $u\in W_{n+1,m+1,\lambda^1}$ then the variety $W_{n,m,\lambda}$ satisfies
the identity
\begin{equation}
\label{u=canon}
u\approx x_1^{\ell_1}\cdots x_t^{\ell_t}x_{t+1}\cdots x_{m+1}\ldotp
\end{equation}

At the rest part of the proof of this lemma, the words `a simple letter' mean
`a simple in $u$ letter'. One can note that one of the following three claims
hold:
\begin{itemize}
\item[1)]the word $u$ ends with a simple letter;
\item[2)]the word $u$ starts with a simple letter;
\item[3)]the word $u$ contains a subword of the kind $x_ix_j$ where $x_i$ and
$x_j$ are simple letters.
\end{itemize}
Indeed, if all these three claims fail then
$$u\equiv w_1y_1w_2y_2\cdots w_{q+1}y_{q+1}w_{q+2}$$
where $y_1,y_2,\dots,y_{q+1}$ are simple letters, while $w_1,w_2,\dots,
w_{q+2}$ are non-empty words such that the word $w\equiv w_1w_2\cdots
w_{q+2}$ does not contain simple letters. Then $r=\ell(w)=\sum
\limits_{i=1}^{q+2}\ell(w_i)\ge q+2$, contradicting the inequality~\eqref
{r<q+2}.

Now we consider three cases corresponding to the claims 1)--3).

\emph{Case} 1: $u\equiv wx_i$ for some word $w$ and some simple letter $x_i$.
The identity
\begin{equation}
\label{w=non-canon}
w\approx x_1^{\ell_1}\cdots x_t^{\ell_t}x_{t+1}\cdots x_{i-1}x_{i+1}\cdots
x_{m+1}
\end{equation}
has the partition $\lambda$, whence it holds in $\mathcal W_{n,m,\lambda}$.
Multiplying~\eqref{w=non-canon} by $x_i$ from the right, we have the identity
\begin{equation}
\label{u=last x_i}
u\approx x_1^{\ell_1}\cdots x_t^{\ell_t}x_{t+1}\cdots x_{i-1}x_{i+1}\cdots
x_{m+1}x_i
\end{equation}
that also holds in $\mathcal W_{n,m,\lambda}$. If $i=m+1$ then~\eqref
{u=last x_i} coincides with~\eqref{u=canon} and we are done. Let now $i\le
m$. Put
$$j=
\begin{cases}
m-1&\text{whenever}\ i=m,\\
m&\text{otherwise}\ldotp
\end{cases}$$
Since $u$ contains at least three simple letters, the letter $x_j$ is simple.
The identity~\eqref{u=last x_i} has the form
\begin{equation}
\label{u=aj,m+1,i}
u\approx ax_jx_{m+1}x_i
\end{equation}
for some $a\in F^1$. Let $x_p$ be a letter with $x_p\notin c(u)$. The
identity
\begin{equation}
\label{ap,i=ai,p}
ax_px_i\approx ax_ix_p
\end{equation}
has the partition $\lambda$, whence it holds in $\mathcal W_{n,m,\lambda}$.
Substituting $x_jx_{m+1}$ for $x_p$ in~\eqref{ap,i=ai,p}, we obtain the
identity
\begin{equation}
\label{aj,m+1,i=ai,j,m+1}
ax_jx_{m+1}x_i\approx ax_ix_jx_{m+1}
\end{equation}
that holds in $\mathcal W_{n,m,\lambda}$. The identity
\begin{equation}
\label{ai,j=short canon}
ax_ix_j\approx x_1^{\ell_1}\cdots x_t^{\ell_t}x_{t+1}\cdots x_m
\end{equation}
has the partition $\lambda$, whence it holds in $\mathcal W_{n,m,\lambda}$
too. Multiplying~\eqref{ai,j=short canon} on $x_{m+1}$ from the right, we
obtain the identity
\begin{equation}
\label{ai,j,m+1=canon}
ax_ix_jx_{m+1}\approx x_1^{\ell_1}\cdots x_t^{\ell_t}x_{t+1}\cdots x_{m+1}
\end{equation}
that holds in $\mathcal W_{n,m,\lambda}$ as well. Combining the
identities~\eqref{u=aj,m+1,i},~\eqref{aj,m+1,i=ai,j,m+1}, and~\eqref
{ai,j,m+1=canon}, we obtain the identity~\eqref{u=canon}.

\emph{Case} 2: $u\equiv x_iw$ for some simple letter $x_i$ and some word $w$.
The word $u$ contains at least three simple letters. Therefore, there is a
simple letter $x_j\in c(w)$. Thus, $w\equiv ax_jb$ for some $a,b\in F^1$. The
identity
\begin{equation}
\label{ajb=abj}
ax_jb\approx abx_j
\end{equation}
has the partition $\lambda$, whence it holds in $\mathcal W_{n,m,\lambda}$.
Multiplying~\eqref{ajb=abj} on $x_i$ from the left, we obtain the identity
$u\approx x_iabx_j$ that also holds in $\mathcal W_{n,m,\lambda}$. We come to
the situation considered in Case~1.

\emph{Case} 3: $u\equiv ax_ix_jb$ for some $a,b\in F^1$ and some simple
letters $x_i$ and $x_j$. Let $x_p$ be a letter with $x_p\notin c(u)$. The
identity
\begin{equation}
\label{apb=abp}
ax_pb\approx abx_p
\end{equation}
has the partition $\lambda$, whence it holds in $\mathcal W_{n,m,\lambda}$.
Substituting $x_ix_j$ for $x_p$ in~\eqref{apb=abp}, we obtain the identity
$u\approx abx_ix_j$ that holds in $\mathcal W_{n,m,\lambda}$. We come to the
situation considered in Case~1 again.
\end{proof}

\begin{corollary}
\label{S_lambda in W_lambda^k}
If $\lambda\in\Lambda$ then $\mathcal{S_\lambda\subseteq W}_{\lambda^k}$ for
any $k\ge0$.
\end{corollary}

\begin{proof}
If $k\le s(\lambda)$ then the desired inclusion holds by the definition of
the variety $\mathcal S_\lambda$. Let now $k>s(\lambda)$. It is easy to see
that $s\bigl(\lambda^{s(\lambda)}\bigr)=0$. Now we may apply Lemma~\ref
{s(lambda)=0} and conclude that $\mathcal{S_\lambda\subseteq
W}_{\lambda^{s(\lambda)}}\subseteq\mathcal W_{\lambda^k}$.
\end{proof}

\begin{proposition}
\label{S_lambda is greedy}
If $\lambda\in\Lambda$ then the variety $\mathcal S_\lambda$ is greedy.
\end{proposition}

\begin{proof}
Suppose that $\mu\in\Lambda$ and the variety $\mathcal S_\lambda$ reduces the
transversal $W_\mu$, that is $\mathcal S_\lambda$ satisfies some non-trivial
identity $u\approx v$ with $u,v\in W_\mu$. Lemma~\ref{lambda less u=v}
implies that
$\lambda\preceq\mu$, that is $\mu=U_S\bigl(\lambda^k\bigr)$ for some $S$ and
$k$. Applying Corollary~\ref{S_lambda in W_lambda^k} and Lemma~\ref
{W_lambda in W_mu}, we have $\mathcal{S_\lambda\subseteq W}_{\lambda^k}
\subseteq\mathcal W_\mu$. Thus, if $s,t\in W_\mu$ then the identity $s\approx
t$ holds in $\mathcal S_\lambda$. This means that $\mathcal S_\lambda$
collapses $W_\mu$. We see that the variety $\mathcal S_\lambda$ collapses any
transversal it reduces, that is $\mathcal S_\lambda$ is greedy.
\end{proof}

Now we are well prepared to prove the `only if' part of Proposition \ref
{reduce}. Let an overcommutative variety $\mathcal V$ satisfy the
condition~(vi) of Theorem~\ref{main}. We need to verify that $\mathcal V$ is
greedy. It is evident that the variety $\mathcal{SEM}$ is greedy because it
does not reduce any transversal. Let now $\mathcal V=\bigwedge\limits_{i=1}^k
\mathcal S_{\lambda_i}$ for some partitions $\lambda_1,\lambda_2,\dots,
\lambda_k$. By Proposition~\ref{S_lambda is greedy}, the varieties $\mathcal
S_{\lambda_1}$, $\mathcal S_{\lambda_2}$, \dots, $\mathcal S_{\lambda_k}$ are
greedy. Proposition~\ref{greedy} implies now that all these varieties are
neutral elements of the lattice \textbf{OC}. It is well known that the set of
all neutral elements of a lattice $L$ forms a sublattice of $L$ (see~\cite
[Theorem~III.2.9]{Gratzer-98}, for instance). Therefore, $\mathcal V$ is a
neutral element of \textbf{OC}. Now we may apply Proposition~\ref{greedy}
again and conclude that $\mathcal V$ is greedy.

\section{Proof of Proposition \ref{reduce}: sufficiency}
\label{suff}

Here we are going to verify that a greedy overcommutative variety satisfies
the condition (vi) of Theorem~\ref{main}, thus completing the proof of
Proposition~\ref{reduce} and therefore, of Theorem~\ref{main}. We start with
a few easy observations.

\begin{lemma}
\label{if collapses then reduces}
If an overcommutative semigroup variety $\mathcal V$ reduces \textup(in
particular, collapses\textup) a transversal $W_\lambda$ then $\mathcal V$
reduces transversals $W_{\lambda^k}$ for all $k\ge0$.
\end{lemma}

\begin{proof}
The case $k=0$ is obvious because $W_{\lambda^0}=W_\lambda$. Let now $k>0$.
Suppose that $\mathcal V$ satisfies a non-trivial identity of the kind $u
\approx v$ with $u,v\in W_\lambda$. Let $y_1,\dots,y_k$ be letters with $y_1,
\dots,y_k\notin c(u)$. The identity $uy_1\cdots y_k\approx vy_1\cdots y_k$ is
non-trivial and holds in $\mathcal V$ because it follows from $u\approx v$.
Since
$$\partition(uy_1\cdots y_k\approx vy_1\cdots y_k)=\lambda^k,$$
$\mathcal V$ reduces $W_{\lambda^k}$.
\end{proof}

\begin{lemma}
\label{V in S_lambda}
Let $\mathcal V$ be a greedy variety. If a non-trivial identity $u\approx v$
holds in $\mathcal V$ and $\partition(u\approx v)=\lambda$ then $\mathcal{V
\subseteq S_\lambda}$.
\end{lemma}

\begin{proof}
By Lemma~\ref{if collapses then reduces}, $\mathcal V$ reduces transversals
$W_{\lambda^k}$ for all $k=0,1,\dots,s(\lambda)$. Being greedy, $\mathcal V$
collapses all these transversals. Therefore, $\mathcal{V\subseteq
S_\lambda}$.
\end{proof}

\begin{corollary}
\label{S_lambda in S_mu}
Let $\lambda,\mu\in\Lambda$. Then $\mathcal{S_\lambda\subseteq S_\mu}$ if and
only if $\lambda\preceq\mu$.
\end{corollary}

\begin{proof}
\emph{Necessity}. Suppose that $\mathcal{S_\lambda\subseteq S_\mu}$. Let $u
\approx v$ be an identity with $\partition(u\approx v)=\mu$. Then $u\approx
v$ holds in $\mathcal S_\mu$, whence it holds in $\mathcal S_\lambda$. Now
Lemma~\ref{lambda less u=v} applies with the conclusion that $\lambda\preceq
\mu$.

\emph{Sufficiency}. Let $\lambda\preceq\mu$ and $u\approx v$ an identity with
$\partition(u\approx v)=\lambda$. Since $\lambda\preceq\mu$, there is an
identity $s\approx t$ such that $u\approx v$ implies $s\approx t$ and
$\partition(s\approx t)=\mu$. The variety $\mathcal S_\lambda$ satisfies the
identity $u\approx v$. Hence $s\approx t$ holds in $\mathcal S_\lambda$ as
well. According to Proposition~\ref{S_lambda is greedy}, the variety
$\mathcal S_\lambda$ is greedy. Now Lemma~\ref{V in S_lambda} succsessfully
applies with the conclusion that $\mathcal{S_\lambda\subseteq S_\mu}$.
\end{proof}

Now we are ready to prove the `if' part of Proposition \ref{reduce}. Let
$\mathcal V$ be a greedy variety and $\mathcal{V\ne SEM}$. The last
unequality means that $\mathcal V$ satisfies some non-trivial identity $u
\approx v$. Put $\lambda=\partition(u\approx v)$. Lemma~\ref{V in S_lambda}
shows that $\mathcal{V\subseteq S_\lambda}$. Thus, the set $\Gamma=\{\lambda
\in\Lambda\mid\mathcal{V\subseteq S_\lambda}\}$ is non-empty. Put $\mathcal
X=\bigwedge\limits_{\lambda\in\Gamma}\mathcal S_\lambda$. Clearly, $\mathcal
{V\subseteq X}$. Suppose that $\mathcal{V\ne X}$. Then there is an identity
$u\approx v$ that holds in $\mathcal V$ but fails in $\mathcal X$. Let $\mu=
\partition(u\approx v)$. By Lemma~\ref{V in S_lambda}, $\mathcal{V\subseteq
S_\mu}$. This means that $\mu\in\Gamma$, whence $\mathcal{X\subseteq S_\mu}$.
Since $\mathcal W_\mu$ satisfies $u\approx v$ and $\mathcal{X\subseteq S_\mu
\subseteq W_\mu}$, we have that the identity $u\approx v$ holds in $\mathcal
X$. A contradiction shows that $\mathcal{V=X}$.

Lemma~\ref{min condition} and Corollary~\ref{S_lambda in S_mu} imply together
that $\mathcal V=\bigwedge\limits_{\lambda\in\Gamma'}\mathcal S_\lambda$
where $\Gamma'$ is the set of all minimal elements of the partially ordered
set $\langle\Gamma;\preceq\rangle$. Since $\Gamma'$ forms an anti-chain in
$\langle\Lambda;\preceq\rangle$, Lemma~\ref{antichains} implies that the set
$\Gamma'$ is finite. Thus, $\mathcal V$ satisfies the condition~(vi) of
Theorem~\ref{main}.

Proposition~\ref{reduce} and Theorem~\ref{main} are proved.\qed

\section{Additional remarks}
\label{optim}

Here we are going to show that the description of the varieties under
consideration given by Theorem~\ref{main} may not be improved, in a sense.
Theorem~\ref{main} shows that the varieties of the kind $\mathcal S_\lambda$
play the crucial role in the description of varieties we consider in this
article. Recall that the variety $\mathcal S_\lambda$ is defined as the
intersection of the varieties $\mathcal W_{\lambda^i}$ where $i$ runs over
the set $\{0,1,\dots,s(\lambda)\}$. A natural question arises, whether or not
the number $s(\lambda)$ may be changed on some lesser number here.

For any $\lambda\in\Lambda$ and $k\in\{0,1,\dots,s(\lambda)\}$, we put
$$\mathcal S_\lambda^k=\bigwedge_{i=0}^k\mathcal W_{\lambda^i}\ldotp$$
In particular, $\mathcal S_\lambda^0=\mathcal W_\lambda$ and $\mathcal
S_\lambda^{s(\lambda)}=\mathcal S_\lambda$. The crucial property of the
variety $\mathcal S_\lambda$ is given by Proposition~\ref
{S_lambda is greedy}: this variety is greedy. The following statement
together with Lemma~\ref{if collapses then reduces} show that varieties
$\mathcal S_\lambda^k$ with $k<s(\lambda)$ does not have this property. Thus,
the question posed in the previous paragraph is answered in negative.

\begin{proposition}
\label{optimum}
Let $\lambda\in\Lambda$, $s(\lambda)>0$, and $0\le k<s(\lambda)$. Then the
variety $\mathcal S_\lambda^k$ does not collapse the transversals
$W_{\lambda^{k+1}}$, $W_{\lambda^{k+2}}$, \dots, $W_{\lambda^{s(\lambda)}}$.
\end{proposition}

\begin{proof}
Let $i\in\{k+1,\dots,s(\lambda)\}$. Suppose that $\mathcal S_\lambda^k$
collapses the transversal $W_{\lambda^i}$. Further considerations are divided
into two cases.

\emph{Case} 1: $\lambda\ne(2,1)$. The definition of the number $s(\lambda)$
and the inequality $s(\lambda)>0$ imply that $s(\lambda)=r(\lambda)-q
(\lambda)-1$ here. Since $s(\lambda)\ge i$, we have $r(\lambda)-q(\lambda)-1
\ge i$. Evident equalities $r\bigl(\lambda^i\bigr)=r(\lambda)$ and $q\bigl
(\lambda^i\bigr)=q(\lambda)+i$ then imply that $r\bigl(\lambda^i\bigr)\ge q
\bigl(\lambda^i\bigr)+1$. Hence the transversal $W_{\lambda^i}$ contains a
word $u$ of the kind
$$u\equiv w_1y_1w_2y_2\cdots w_{q+i}y_{q+i}w_{q+i+1}$$
where $y_1,y_2,\dots,y_{q+i}$ are simple in $u$ letters, while $w_1,w_2,
\dots,w_{q+i+1}$ are non-empty words such that the word $w_1w_2\cdots
w_{q+i+1}$ does not contain simple in $u$ letters. Let $v\in W_{\lambda^i}$
and $v\not\equiv u$. Since $\mathcal S_\lambda^k$ collapses $W_{\lambda^i}$,
the identity $u\approx v$ holds in $S_\lambda^k$. Therefore, this identity
follows from the identity system
$$\Sigma=\{g\approx h\mid\ \text{there is}\ j\in\{0,1,\dots,k\}\ \text{with}\
g,h\in W_{\lambda^j}\}\ldotp$$
Let~\eqref{deduct} be a deduction of $u\approx v$ from $\Sigma$. We have $u
\equiv a\zeta(s)b$ and $w_1\equiv a\zeta(t)b$ for some homomorphism $\zeta$
on $F$, some $a,b\in F^1$, and some $s,t\in W_{\lambda^j}$ where $j\in\{0,1,
\dots,k\}$. Furthermore,
\begin{equation}
\label{equalities}
r(\partition(s))=r\bigl(\lambda^j\bigr)=r\bigl(\lambda^i\bigr)=r(\partition
(u))=r(\partition(a\zeta(s)b))\ldotp
\end{equation}
On the other hand, it is evident that
\begin{equation}
\label{unequalities}
r(\partition(s))\le r(\partition(\zeta(s)))\le r(\partition(a\zeta(s)b))
\ldotp
\end{equation}
Combining~\eqref{equalities} and~\eqref{unequalities}, we have
\begin{equation}
\label{one more equalities}
r(\partition(s))=r(\partition(\zeta(s)))=r(\partition(a\zeta(s)b))\ldotp
\end{equation}
This implies that the subword $\zeta(s)$ of the word $a\zeta(s)b$ contains
all occurrences of non-simple in $a\zeta(s)b$ letters, whence all letters
from $c(ab)$ are simple in $u$. But the word $a\zeta(s)b\equiv u$ starts and
ends with non-simple in $u$ letters. Therefore, the words $a$ and $b$ are
empty. Thus, $u\equiv\zeta(s)$. If either there is a non-simple in $s$ letter
$x$ with $\ell(\zeta(x))>1$ or there is a simple in $s$ letter $y$ such that
the word $\zeta(y)$ contains some non-simple in $u$ letter then $r(\partition
(\zeta(s)))>r(\partition(s))$, contradicting~\eqref{one more equalities}.
Hence $\zeta$ maps every non-simple in $s$ letter to a letter and maps every
simple in $s$ letter to a word consisting of simple letters. If $y$ is a
simple in $s$ letter then $\zeta(y)$ is a subword of $u$. But $u$ does not
contain subwords consisting of simple letters except subwords of length~1,
that is letters. Thus, $\zeta$ maps a simple in $s$ letter to a simple in
$\zeta(s)\equiv u$ letter. We conclude that $\ell(u)=\ell(\zeta(s))=\ell(s)$.
But $\ell(s)=n+j$ and $\ell(u)=n+i$ where $n=\ell(\lambda)$. Therefore, $j=
i$. But this is impossible because $i\ge k+1$, while $j\le k$.

\emph{Case} 2: $\lambda=(2,1)$. Here $r(\lambda)=2$, $q(\lambda)=1$, and
$\delta=0$, whence $s(\lambda)=1$. Therefore, $k=0$ and $i=1$. This means
that $\mathcal S_\lambda^k=\mathcal W_\lambda$ and $\lambda^i=(2,1,1)$. Let
$u\equiv x^2yz$ and $v\equiv x^2zy$. Suppose that the identity $u\approx v$
holds in $\mathcal W_\lambda$. Then it follows from the identity system
$$\Sigma=\{x^2y\approx xyx\approx yx^2\}\ldotp$$
Let~\eqref{deduct} be a deduction of $u\approx v$ from $\Sigma$. Then there
is $j\in\{0,1,\dots,\ell\}$ such that the first occurrence of $z$ in the word
$w_j$ precedes the first occurrence of $y$ in $w_j$. Let $j$ be the least
number with such a property. It is evident that $j>0$. Thus, the following
holds:
\begin{align}
\label{w_j-1}&w_{j-1}\in\{x^2yz,xyxz,xyzx,yx^2z,yxzx,yzx^2\},\\
\label{w_j}&w_j\in\{x^2zy,xzxy,xzyx,zx^2y,zxyx,zyx^2\}\ldotp
\end{align}
Furthermore, $w_{j-1}\equiv a\zeta(s)b$ and $w_j\equiv a\zeta(t)b$ for some
homomorphism $\zeta$ on $F$, some $a,b\in F^1$, and some $s,t\in\{x^2y,xyx,y
x^2\}$. Repeating mutatis mutandi arguments from Case~1), we obtain that $r
(\partition(w_{j-1}))=r(\partition(w_j))=2$ and deduct from these equalities
that $x\notin c(ab)$, $\zeta(x)$ is a letter, and $\zeta(y)\in\{y,z,yz,zy\}$.
If $\zeta(x)\equiv e$ then $e$ is a non-simple in $w_{j-1}$ letter. In view
of~\eqref{w_j-1}, $\zeta(x)\equiv x$.

If $\zeta(y)\equiv yz$ then the word $w_j$ contains the subword $yz$,
contradicting~\eqref{w_j}. Analogously, if $\zeta(y)\equiv zy$ then the word
$w_{j-1}$ contains the subword $zy$, contradicting~\eqref{w_j-1}.

Suppose now that $\zeta(y)\equiv y$. Then $\zeta(s)\equiv s$ and $\zeta(t)
\equiv t$. Since $\zeta(s)$ is a subword in $w_{j-1}$, this means that one of
the words $x^2y$, $xyx$ or $yx^2$ is a subword in $w_{j-1}$. In view
of~\eqref{w_j-1}, this means that $w_{j-1}$ coincides with one of the words
$x^2yz$, $xyxz$ or $yx^2z$. Thus, the word $a$ is empty and therefore, $w_j
\equiv\zeta(t)b\equiv tb$. Since $z\notin c(t)$, we have that the first
occurrence of $y$ in $w_j$ precedes the first occurrence of $z$ in $w_j$. But
this contradicts the choice of the number $j$.

Finally, let $\zeta(y)\equiv z$.  Then $\zeta(s)\in\{x^2z,xzx,zx^2\}$, whence
$w_{j-1}$ coincides with one of the words $ax^2zb$, $axzxb$ or $azx^2b$. In
view of~\eqref{w_j-1}, this means that $w_{j-1}\in\{yx^2z,yxzx,yzx^2\}$.
Therefore, $a\equiv y$. Thus, the word $w_j$ starts with the letter $y$. As
in the previous paragraph, we have that the first occurrence of $y$ in $w_j$
precedes the first occurrence of $z$ in $w_j$ that contradicts the choice of
$j$.

We prove that the variety $\mathcal{W_\lambda=S}_\lambda^k$ does not satisfy
the identity $x^2yz\approx x^2zy$. Since $\partition(x^2yz\approx x^2zy)=(2,
1,1)=\lambda^i$, we have that $\mathcal S_\lambda^k$ does not collapse the
transversal $W_{\lambda^i}$.
\end{proof}

One can return to the definition of the variety $\mathcal S_\lambda$. It may
be written in the form
\begin{equation}
\label{definition of S__lambda}
\mathcal S_\lambda=\bigwedge_{\mu\in\Gamma}\mathcal W_\mu
\end{equation}
where $\Gamma=\{\lambda^k\mid k=0,1,\dots,s(\lambda)\}$. The following
assertion shows that the set $\{\lambda^k\mid k=0,1,\dots,s(\lambda)\}$ is
the least set of partitions $\Gamma$ such that the equality~\eqref
{definition of S__lambda} holds.

\begin{corollary}
\label{representation of S_lambda}
If the equality~\textup{\eqref{definition of S__lambda}} holds for some
$\Gamma\subseteq\Lambda$ then $\lambda^k\in\Gamma$ for all $k=0,1,\dots,s
(\lambda)$.
\end{corollary}

\begin{proof}
Suppose that $\lambda^k\notin\Gamma$ for some $k\in\{0,1,\dots,s(\lambda)\}$.
Let $u,v\in W_{\lambda^k}$. The definition of the variety
$\mathcal S_\lambda$ implies that the identity $u\approx v$ holds in
$\mathcal S_\lambda$. Therefore, this identity follows from the identity
system
$$\Sigma=\{g\approx h\mid\text{there is}\ \mu\in\Gamma\ \text{such that}\ g,h
\in W_\mu\}\ldotp$$
As usual, let~\eqref{deduct} be a deduction of $u\approx v$ from $\Sigma$.
Let $i\in\{0,1,\dots,\ell-1\}$. Then the identity $w_i\approx w_{i+1}$
follows from an identity of the kind $s\approx t$ where $s,t\in W_\mu$ for
some $\mu\in\Gamma$. The identity $s\approx t$ holds in the variety $\mathcal
S_\mu$. Therefore, $u\approx v$ holds in $\mathcal S_\mu$ too. Then
Lemma~\ref{lambda less u=v} implies that $\mu\preceq\partition(u\approx v)=
\lambda^k$. Furthemore, the identity $s\approx t$ holds in the variety
$\mathcal S_\lambda$ because $\mathcal{S_\lambda\subseteq W_\mu}$. Applying
Lemma~\ref{lambda less u=v} again, we have $\lambda\preceq\mu$. Therefore,
$\mu=U_S\bigl(\lambda^j\bigr)$ and $\lambda^k=U_T\bigl(\mu^q\bigr)$ for some
finite (may be empty) sets of partitions $S$ and $T$ and some non-negative
integers $j$ and $q$.

Let $n(\lambda)=n$. Then $n(\mu)=n+j$, while $n\bigl(\lambda^k\bigr)$ equals
both $n+k$ (that is evident) and $n+j+q$ (because $n\bigl(\lambda^k\bigr)=n
(\mu)+q$). Therefore, $n+k=n+j+q$, whence $q=k-j$. Thus, $\mu=U_S\bigl
(\lambda^j\bigr)$ and $\lambda^k=U_T\bigl(\mu^{k-j}\bigr)$. Hence $\lambda^k=
U_{S\cup T}\bigl(\lambda^k\bigr)$ and therefore, $S=T=\varnothing$. In
particular, this means that $\mu=U_\varnothing\bigl(\lambda^j\bigr)=
\lambda^j$. We see that $\lambda^j=\mu\preceq\lambda^k$, whence $j\le k$. But
$j\ne k$ because $\lambda^k\notin\Gamma$, while $\lambda^j=\mu\in\Gamma$.
Hence $j<k$. Besides that, the equality $\mu=\lambda^j$ implies that the
identity $w_i\approx w_{i+1}$ holds in the variety $\mathcal W_{\lambda^j}$.
Since $j\le k-1$, we have $\mathcal S_\lambda^{k-1}\subseteq\mathcal
W_{\lambda^j}$. Thus, $w_i\approx w_{i+1}$ holds in $\mathcal
S_\lambda^{k-1}$. This is the case for all $i=0,1,\dots,\ell-1$. Therefore,
the identity $u\approx v$ holds in $\mathcal S_\lambda^{k-1}$ too. This is
valid for all $u,v\in W_{\lambda^k}$. Hence the variety $\mathcal
S_\lambda^{k-1}$ collapses the transversal $W_{\lambda^k}$, contradicting
Proposition~\ref{optimum}.
\end{proof}

\section{Open problems}
\label{quest}

Recall that an element $x$ of a lattice $L$ is called \emph{modular} if
$$\forall y,z\in L\colon\quad y\le z\longrightarrow(x\vee y)\wedge z=(x\wedge
z)\vee y,$$
and \emph{upper-modular} if
$$\forall y,z\in L\colon\quad y\le x\longrightarrow x\wedge(y\vee z)=(x\wedge
z)\vee y\ldotp$$
\emph{Lower-modular} elements are defined dually to upper-modular ones.

\begin{problem}
\label{mod in OC}
Describe
\begin{itemize}
\item[\textup{a)}]modular;
\item[\textup{b)}]upper-modular;
\item[\textup{c)}]lower-modular
\end{itemize}
elements of the lattice $\mathbf{OC}$.
\end{problem}

As we have already mentioned in Section~\ref{intr}, neutral elements of the
lattice \textbf{SEM} are completely determined in~\cite{Volkov-05}, while
distributive elements of this lattice are completely described in~\cite
{Vernikov-Shaprynskii-distr}. It is interesting to note that Proposition~\ref
{greedy} plays an important role in the proof of the result of~\cite
{Vernikov-Shaprynskii-distr}.

\begin{problem}
\label{distr in SEM}
Describe
\begin{itemize}
\item[\textup{a)}]codistributive;
\item[\textup{b)}]standard;
\item[\textup{c)}]costandard
\end{itemize}
elements of the lattice $\mathbf{SEM}$.
\end{problem}

Some particular results concerning Problem~\ref{distr in SEM}a) are obtained
in~\cite{Vernikov-codistr}. The following two examples show that, in contrast
with the overcommutative case, the lattice \textbf{SEM} contains distributive
but not codistributive elements and codistributive but not distributive ones.
In particular, there are [co]distributive but not neutral elements of \textbf
{SEM}.

\begin{example}
\label{distr not codistr}
The variety $\mathcal N=\var\{x^2y=xyx=yx^2=0\}$ is a distributive element of
the lattice \textbf{SEM} by~\cite[Theorem~1.1]{Vernikov-Shaprynskii-distr}.
But this variety is not a codistributive (and moreover not a neutral) element
of \textbf{SEM} by~\cite[Theorem~1.1]{Vernikov-codistr}.
\end{example}

\begin{example}
\label{codistr not distr}
The varieties $\mathcal A_p=\var\{x^py=y,xy=yx\}$ with any prime $p$,
$\mathcal{LZ}=\var\{xy=x\}$, and $\mathcal{RZ}=\var\{xy=y\}$ are
codistributive elements of the lattice \textbf{SEM}. This follows from the
well known facts that these varieties are atoms of \textbf{SEM} and \textbf
{SEM} satisfies the condition
$$\forall x,y,z\colon\quad x\wedge z=y\wedge z=0\longrightarrow(x\vee y)
\wedge z=0$$
(see~\cite[Section~1]{Shevrin-Vernikov-Volkov-09}, for instance). But
$\mathcal A_p$, $\mathcal{LZ}$, and $\mathcal{RZ}$ are not distributive (and
moreover not neutral) elements of \textbf{SEM} by~\cite[Theorem~1.1]
{Vernikov-Shaprynskii-distr}.
\end{example}

\small

redpajama_set_name: RedPajamaArXiv


The Cohn-Kanade AU-Coded Facial Expression Database is for research in automatic facial image analysis and synthesis and for perceptual studies. Cohn-Kanade is available in two versions and a third is in preparation. Version 1 (the original or initial release (Kanade, Cohn, & Tian, 2000)) includes 486 sequences from 97 posers. Each sequence begins with a neutral expression and proceeds to a peak expression. The peak expression for each sequence is fully FACS (Ekman, Friesen, & Hager, 2002; Ekman & Friesen, 1979) coded and given an emotion label. The emotion label refers to what expression was requested rather than what may actually have been performed. For validated emotion labels, please use version 2, CK+, as described below. Version 2, referred to as CK+, includes both posed and non-posed (spontaneous) expressions and additional types of metadata. For posed expressions, the number of sequences is increased from the initial release by 22% and the number of subjects by 27%. As with the initial release, the target expression for each sequence is fully FACS coded. In addition validated emotion labels have been added to the metadata. Thus, sequences may be analyzed for both action units and prototypic emotions. The non-posed expressions are from Ambadar, Cohn, & Reed (2009). Additionally, CK+ provides protocols and baseline results for facial feature tracking and action unit and emotion recognition. Tracking results for shape and appearance are via the approach of Matthews & Baker (2004). For action unit and expression recognition, a linear support vector machine (SVM) classifier with leave-one-out subject cross-validation was used. Both sets of results are included with the metadata. For a full description of CK+, please see P. Lucey et al. (2010). Version 3 is planned for spring 2013. The original data collection of Cohn-Kanade included synchronized frontal and 30-degree from frontal video (fig. 1, below). Version 3 will add the synchronized 30-degree from frontal video. To receive the database for research, non-commercial use, download, sign, and return an Agreement to the Affect Analysis Group. All student or non-faculty agreement forms must be co-signed by a faculty advisor.


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ONF unveils global SD-RAN project
Written by Scott Bicheno
A bunch of operators and tech players have come together to try to accelerate the development of open, software-driven radio access network technology.
There are already a number of initiatives in this area, most conspicuously O-RAN, but it seems some operators aren't happy with the pace of progress and have decided to take matters in hand. Founding members include AT&T, China Mobile and Deutsche Telekom as well as US tech giants Facebook, Google and Intel.
The initial focus of the project will be on building an open source Near Real-Time RAN Intelligent Controller (nRT-RIC) compatible with the O-RAN architecture. So, while the very creation of this group suggests some degree of dissatisfaction with the efforts of the O-RAN group, there's the implicit acknowledgement that any incompatibility would totally undermine the 'open' ethos. The presence of Facebook shows that diplomatic relations with TIP are strong too.
"AT&T strongly supports the development of specifications and components that can help drive openness and innovation in the RAN ecosystem," said Andre Fuetsch, CTO of AT&T Labs. "The O-RAN Alliance's specifications are enabling the ecosystem, with a range of companies and organizations creating both open source and proprietary implementations that are bringing the open specifications to life… This project will help accelerate the transition to an open RAN future."
"China Mobile co-founded O-RAN in order to promote both the opening of the RAN ecosystem for multi-vendor solutions and the realization of RAN with native intelligence for performance and cost improvement," said Dr. Chih-Lin I, Chief Scientist, Wireless Technologies, China Mobile. "We are excited to see the development of an open nRT-RIC and xApps in the SD-RAN project led by ONF, and expect this work to help advance the state-of-art for open and intelligent RAN."
"Deutsche Telekom is a huge believer in applying disaggregation and open source principles for our next-generation networks," said Dr. Alex Jinsung Choi, SVP Strategy & Technology Innovation, Deutsche Telekom. "DT has ONF's mobile core platform (OMEC) in production and we are taking ONF's broadband access (SEBA/VOLTHA) platform to production towards the end of 2020.
"This journey has shown us the tremendous value that is created when we can build solutions based on interoperable multi-vendor components intermixed with open source components. ONF's SD-RAN project is leveraging these same principles to help accelerate innovation in the RAN domain, and we are excited to be an active collaborator in this journey."
So everyone's making all the right noises about collaboration and interoperability, but it's still not obvious why that couldn't have been done under the auspices of existing open RAN groups. The absence of any of the big kit vendors, while not surprising, is also noteworthy as this kind of technology is generally viewed as an existential threat to them.
Tags: AT&T, China Mobile DT, featured, Open RAN
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